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4

The product rule tells us $$\nabla |\nabla u| = \frac{\nabla (g(\nabla u,\nabla u))}{2 |\nabla u|}=\frac{g(\nabla^2u,\nabla u)}{|\nabla u|}.$$ Taking the squared norm and estimating the numerator with Cauchy-Schwarz gives us $$|\nabla|\nabla u||^2 \le \frac{|\nabla^2 u|^2 |\nabla u|^2}{|\nabla u|^2} = |\nabla^2 u|^2$$ as desired.


3

This is studied in potential theory: the function $u$ is the Newtonian potential of $f$, $$u(x)=\int_{\mathbb{R}^n} K(x-y)f(y)\,dy$$ where $K(x)=c_n|x|^{2-n}$ for $n\ne 2$ and $K(x)=c_2\log|x|$ for $n=2$. In dimensions $n\ge 3$ the kernel $K$ decays at infinity, so $u(x)\to 0$ as $|x|\to\infty$ in this case, provided $f$ is reasonable (integrable and ...


3

It's primarily due to the similarity to the corresponding algebraic equations. For example, let $A \in \mathbb{R}^{n\times n}$ be a symmetric positive definite matrix. We say that the PDEs: $$ \begin{cases} \sum_{i,j=1}^n A_{ij} \partial_i \partial_j u =f &\text{is elliptic} \\ \partial_t u - \sum_{i,j=1}^n A_{ij} \partial_i \partial_j u ...


2

It depends on your definition of $\|f\|_{H^{-1}}$. Let us equip $H_0^1$ with the scalar product $$(u,v)_{H_0^1} = \int \nabla u \nabla v \, \mathrm dx.$$ Then, the weak formulation of Poisson's equation is $$(\phi,v)_{H_0^1} = f(v) \quad\forall v \in H_0^1.$$ Hence, the solution $\phi \in H_0^1$ is just the Riesz representative of $f \in H^{-1} = (H_0^1)'$. ...


2

The Spectral Theorem for $A$ is given in terms of a Borel Spectral measure $E$ $$ Ax = \int_{-\infty}^{\infty}\lambda dE(\lambda)x, $$ and $x \in \mathcal{D}(A)$ iff $$ \int_{-\infty}^{\infty}\lambda^2 d\|E(\lambda)x\|^2 < \infty. $$ The operator $e^{iA^2}$ is defined through the functional calculus as $$ e^{iA^2}x = ...


1

Anytime you see a finite domain, standard heat equation, think separation of variables. Try to use $u(x,y,t) = T(t)X(x)Y(y),$ and you'll get 3 ODE's to solve, something along the lines of \begin{align} T'(t) + c^2(\lambda_x + \lambda_y)T(t) &= 0\\ X'' - \lambda_x X &=0\\ Y'' - \lambda_y Y &=0. \end{align} This should be the motivated approach ...


1

Unfortunately this is not so simple/immediate and does depend on the type of equation and perturbation. I suggest that before anything else you have a look at the Engel-Nagel book, more precisely at Chapter 3 (such as at their Theorem 3.14). They also include discussions of some specific types of equations, although I don't remind elliptic.


1

Yes. A corollary to the Rellich Kondrachov theorem says that for all $p\ge 1$, we have $W^{1,p}$ compactly contained in $L^p$. Evans PDE book has a nice treatment of this. EDIT: Actually, if I recall correctly, for $\Omega \subseteq \mathbb R^n$, to prove that $W^{1,p}(\Omega)$ is compactly contained in $L^p(\Omega)$, we only need the Rellich Kondrachov ...


1

Setting $v=u_1-u_2$, define $I(t)=\frac{1}{2}\int_0^1v(x,t)^2dx$. Then $$\frac{dI}{dt}=\int_0^1v(x,t)v_t(x,t)dx=\int_0^1vv_{xx}dx$$ Integrating by parts, this becomes: $$\frac{dI}{dt}=\left[vv_x\right]_0^1-\int_0^1v_x^2dx$$ Now, note: $vv_x\vert_{x=0}=0,vv_x\vert_{x=1}=-v(1,t)^2$ by the conditions given, so: ...


1

Assume that $f$ is continuous and that $\phi$, $g$ are differentiable. Suppose $$H(x) = \int_0^x f(s) \, ds.$$ The fundamental theorem of calculus tells you that $H'(x) = f(x)$. Suppose that $$M(x) = \int_0^{\phi(x)} f(s) \, ds.$$ Then $M(x) = H(\phi(x))$ so that the chain rule tells you $$M'(x) = H'(\phi(x)) \phi'(x) = f(\phi(x)) \phi'(x).$$ The ...


1

When considering a function $f(t,\mathbf x)$ on $[0,\infty)\times \mathbb{R}^n$, we can focus on one time slice at a time, fixing $t$ and dealing with a function of $\mathbf x$ only. The Fourier transform can be applied to this slice, since it's just a function on $\mathbb{R}^n$. The result can be denoted $\tilde f(t,\mathbf k)$, and is usually called the ...


1

Apart from a few typos, your derivation looks correct. There should be $e^{-\alpha^2 t}$ in the final equation and I get $x + \tan(x) = 0$ (which is $\cot(x) = -\frac{1}{x}$), as the defining equation for the $\alpha_n$'s. This follows from $$X(x) = \sin(\alpha x) \implies 0 = X(1) + X'(1) = \sin(\alpha) + \alpha \cos(\alpha) \implies \tan(\alpha) + \alpha ...


1

Using the Spectral Theorem, $$ Ax=\int_{0}^{\infty}\lambda dE(\lambda)x \\ \mathcal{D}(A) = \left\{ x : \int_{0}^{\infty}\lambda^2 d\|E(\lambda)x\|^2 < \infty \right\}. $$ Then the positive square root $\sqrt{A}$ is $$ \sqrt{A}x = \int_{0}^{\infty}\sqrt{\lambda}dE(\lambda)x \\ \mathcal{D}(\sqrt{A}) = \left\{ x : ...


1

Hint: Since you have a final condition $v(T,x)=x^2$ instead of an initial condition, and you would like to solve the problem backwards, for the time interval $t \in (0,T)$, it is very useful to reverse time, that is, to introduce $\tau = T- t$, such that the equation becomes \begin{equation} v_{\tau} = \frac{1}{2} v_{xx},\qquad v(\tau=0,x) = x^2 ...


1

Your approach is perfectly appropriate. There was, however, a flaw in the execution of that way forward. Note that we have $$\begin{align} F(k)&=\langle H(a-|x|),e^{-ikx}\rangle\\\\ &=\frac{1}{-ik}\langle H(a-|x|),\frac{de^{-ikx}}{dx}\rangle\\\\ &=\frac{1}{-ik}\langle \text{sgn}(x)H'(a-|x|),e^{-ikx})\\\\ &=-\frac{1}{ik}\langle ...


1

$$u_{xy}-xu_x+u=0$$ This elliptic PDE is already on canonical form. Seach for particular solutions. Method of separation of variables, with $u=f(x)g(y)$ $$f'g'-xf'g+fg=0$$ $\frac{g'}{g}=x-\frac{1}{\frac{f'}{f}}=$(function of $y$)=(function of $x$)=constant=$\lambda$ $$\begin{cases} \frac{g'}{g}=\lambda \\ (x-\lambda)f'-f=0 \end{cases} \qquad\to\qquad ...


1

I think you have misunderstood the nature of the solutions to Bessel's equation: the differential equation $$ y'' + \frac{1}{x}y' + \left(\lambda^2-\frac{n^2}{x^2}\right) y = 0 $$ (Bessel's equation, with eigenvalue $\lambda^2$, $\lambda>0$) has a regular singular point at $x=0$, an irregular singular point at $\infty$, and all other points are regular ...


1

With respect to the new bounty: Here's a way to see this. Suppose $u$ is a compactly supported solution to $\bar{\partial}u=f$. Then we have, for large enough $R>0$ $$0=\int_{\partial D(0,R)} u(z)dz = 2i\int_{D(0,R)} \bar{\partial}u(z)\: d\bar{z}\wedge dz=2i\int_{D(0,R)} f(z)\:d\bar{z}\wedge dz$$ Taking $R\to\infty$, this implies that if a compactly ...


1

Try rechecking your work. The derivative of $\ln(f(x))$ is $f'/f$ . Observe $$\frac{\partial w}{\partial x}= \frac{2 u \frac{\partial u }{\partial x }+2v\frac{\partial v }{\partial x} } {u^2 + v^2} $$ where $\frac{\partial u }{\partial x} = e^{x^2 +y} (2x)$ and $\frac{\partial v}{\partial x} = e^{x + y^2}$ When $(x,y) = 0$ we have that $u = 1$, $v = ...


1

A(t) can be solved from the first equation: $\frac{du}{dt} = k \frac{d^2u}{dx^2} \\ A'(t)\sin\frac{\pi x}{L} = -k \frac{\pi^2}{L^2} A(t) \sin\frac{\pi x}{L} \\A'(t)= -k \frac{\pi^2}{L^2} A(t) \\ A(t) = c_1 \exp(-k\frac{\pi^2}{L^2} t)$ and $c$ cannot be solved without an initial condition. This can be substituted into the two equation for solving for the B ...


1

This seems to work: The hyperbolicity implies $$ \theta \int_\Omega |Du|^2 \, dx \leq \int \sum_{i,j} a^{ij} u_i u_j \, dx = B[u,u] - \int_\Omega c u^2 \, dx \leq B[u,u]\,, $$ since $c \geq 0$. For $u \in H_0^1(\Omega)$ the Poincaré inequality implies that there exists $C \geq 0$ such that $$ \|u\|_{H^1(\Omega)}^2 = \|u\|_{L^2(\Omega)}^2 + ...



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